# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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then

Y'(t) = eat [u'(t) + au (t)]

and

Y"(t) = eat [u"(t) + 2au'(t) + a2u (t)].

Substituting for y, y, and Ó in Eq. (27), canceling the factor eat, and collecting terms, we obtain

au"(t) + (2aa + b)u'(t) + (aa2 + ba + c)u (t) = Pn (t). (28)

The determination of a particular solution ofEq. (28) is precisely the same problem, except for the names of the constants, as solving Eq. (24). Therefore, if aa2 + ba + c is not zero, we assume that u(t) = A0tn + ... + An; hence a particular solution of Eq. (27) is of the form

Y (t) = eat (A0tn + A1tn-1 + ... + An). (29)

On the other hand, if aa2 + ba + c is zero, but 2aa + b is not, we must take u(t) to be of the form t(A0tn + ... + An). The corresponding form for Y(t) is t times the expression on the right side ofEq. (29). Note that if aa2 + ba + c is zero, then eat is a solution of the homogeneous equation. If both aa2 + ba + c and 2aa + b are zero (and this implies that both eat and teat are solutions of the homogeneous equation), then the correct form for u(t) is t2( A0tn + ... + An). Hence Y(t) is t2 times the expression on the right side ofEq. (29).

g(t) = eatPn(t) cos â t or eatPn(t) sin â t. These two cases are similar, so we consider only the latter. We can reduce this problem to the preceding one by noting that, as a consequence of Euler’s formula, sin â t = (e‘et - e~‘et )/2i. Hence g(t) is ofthe form

e(a+ie)t e(a-ie)t

g(t) = Pn (t)-----------2i--------

and we should choose

Y(t) = e(a+ie)t (A0tn + ... + An) + e(a-ie)t (B0tn + ... + Bn), or equivalently,

Y(t) = eat(A0tn + ... + An) cos ât + eat(B0tn + ... + Bn) sin â t.

Usually, the latter form is preferred. If a ± ³â satisfy the characteristic equation corresponding to the homogeneous equation, we must, of course, multiply each of the polynomials by t to increase their degrees by one.

If the nonhomogeneous function involves both cos â t and sin â t, it is usually convenient to treat these terms together, since each one individually may give rise to the

178

Chapter 3. Second Order Linear Equations

same form for a particular solution. For example, if g(t) = t sin t + 2cos t, the form for Y(t) would be

Y(t) = (A0t + A.) sin t + (B0t + Bx) cos t, provided that sin t and cos t are not solutions of the homogeneous equation.

PROBLEMS

In each of Problems 1 through 12 find the general solution of the given differential equation.

1. /- 2/- 3y = 3e^

3. /- 2/- 3y = —3te-t

5. / + 9y = t2e3t + 6 7. 2/' + 3/ + y = t2 + 3 sin t

9. u" + o^u = cos ot, î2 = «2

11.

12.

Ó' + y + 4 y = 2 sinh t y" - Ó - 2y = cosh2t

2. Ó + 2 Ó + 5 y = 3 sin 2t

4. Ó' + 2 Ó = 3 + 4sin2t

6. Ó' + 2 y + y = 2e-t

8. Ó' + y = 3 sin2t + t cos2t

10. è11 + o'2 u = cos o0t

(e‘ - å- ' )/2

(ª + å -t )/2

In each of Problems 13 through 18 find the solution of the given initial value problem.

13. / + /- 2y = 2t, y(0) = 0, Ó (0) = 1

14. y" + 4y = t2 + 3e', y(0) = 0, Ó(0) = 2

15. /- 2/ + y = tet + 4, y(0) = 1, Ó (0) = 1

16. Ó- 2/- 3 y = 3te2t, y(0) = 1, Ó(0) = 0

17. Ó + 4y = 3 sin2t, y(0) = 2, /(0) = -1

18. /' + 2/ + 5y = 4e-tcos2t, y(0) = 1, /(0) = 0

In each of Problems 19 through 26:

(a) Determine a suitable form for Y(t) if the method of undetermined coefficients is to be used.

(b) Use a computer algebra system to find a particular solution of the given equation.

>

>

>

>

>

>

>

19.

21.

22. y

23. /

24. /

25. y

26. Ó

+ 3y = 2t4 + t2e-3t + sin3t > 20. Ó + y = t(1 + sint)

- 5y + 6y = e‘ cos2t + e2(3t + 4) sin t

+ 2y + 2y = 3e-t + 2e-t cos t + 4e-t^ sin t

- 4Ó + 4y = 2^ + 4te2t + tsin2t + 4y = t2 sin2t + (6t + 7) cos2t

+ 3y + 2y = º'(^ + 1) sin2t + 3e-t cos t + 4e‘

+ 2Ó + 5y = 3te-t cos2t - 2te-2t cos t

27. Determine the general solution of

Ó + ë2 y = J]

sin mn t,

m=1

where ê > 0 and Ë = mn for m = 1N.

> 28. In many physical problems the nonhomogeneous term may be specified by different formulas in different time periods. As an example, determine the solution y = ô(³) of

/ + y =

t,

ï ªÏ-t,

0 < t < ï, t > ï,

satisfying the initial conditions y(0) = 0 and Ó (0) = 1. Assume that y and y are also continuous at t = ï. Plot the nonhomogeneous term and the solution as functions of time. Hint: First solve the initial value problem for t < ï; then solve for t > ï, determining the constants in the latter solution from the continuity conditions at t = ï.

3.7 Variation of Parameters

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